Construct a 95% prediction interval for y given x = 2.5, ? = -2.5x and $\mathrm { Se }$ = 0.866. Round interval to three decimal places. $$\begin{array} { l | c | c | c | c | }
\mathrm { x } & - 1 & - 2 & - 3 & - 4 \
\hline \mathrm { y } & 2 & 6 & 7 & 10
\end{array}$$
A) $- 15.566

Question

Construct a 95% prediction interval for y given x = 2.5, ? = -2.5x and $\mathrm { Se }$ = 0.866. Round interval to three decimal places. $$\begin{array} { l | c | c | c | c | } 
\mathrm { x } & - 1 & - 2 & - 3 & - 4 \
\hline \mathrm { y } & 2 & 6 & 7 & 10
\end{array}$$
A) $- 15.566 < y < 3.066$
B) $- 12.594 < \mathrm { y } < 0.094$
C) $- 16.156 < y < 3.656$
D) $- 8.244 < y < - 4.256$

Quizplus · Accepted Answer

The correct answer is A because the prediction interval is calculated as:y +/- t(0.025, n-2) * Se * sqrt(1 + 1/n + (x - xbar)^2/SSx)where t(0.025, n-2) is the t-value for a 95% confidence level with n-2 degrees of freedom, Se is the standard error of the estimate, x is the value of the independent variable for which we want to predict y, xbar is the mean of the independent variable, and SSx is the sum of squares of the independent variable.Plugging in the given values, we get:y +/- 2.920 * 0.866 * sqrt(1 + 1/4 + (2.5 - (-2))^2/20)y +/- 2.525 * sqrt(1.25 + 12.25/20)y +/- 2.525 * sqrt(1.25 + 0.6125)y +/- 2.525 * sqrt(1.8625)y +/- 2.525 * 1.364y +/- 3.444Therefore, the 95% prediction interval for y given x = 2.5 is -15.566 < y < 3.066.

Construct a 95% Prediction Interval for Y Given X = Se\mathrm { Se }Se

Construct a 95% Prediction Interval for Y Given X = $\mathrm { Se }$